What is the sum of all positive integers smaller than 1000 that can be written in the form 100x2^n, where n is an integer (not necessarily positive)?

Question

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Answer 1

100×2ⁿ=25×4×2ⁿ=25×2ⁿ⁺².

Let p and q be two positive integers. The sum of all the positive integers up to each of these is p(p+1)/2 and q(q+1)/2. That is, the sum of all integer x for q<x≤p. The difference between these is the sum of all the positive integers between p and q. So if p>q, p(p+1)/2-q(q+1)/2=25×2ⁿ⁺², p²+p-q²-q=25×2ⁿ⁺³, p²-q²+p-q=25×2ⁿ⁺³, (p-q)(p+q)+(p-q)=25×2ⁿ⁺³, (p-q)(p+q+1)=25×2ⁿ⁺³.

So, let p+q+1=25, therefore p+q=24; and p-q=2ⁿ⁺³.

Add these two equations: 2p=24+2ⁿ⁺³, p=12+2ⁿ⁺² and q=24-p=12-2ⁿ⁺².

Since q>0, 12-2ⁿ⁺²>0, 3-2ⁿ>0, 2ⁿ<3, so n≤1. Because p and q are positive integers, n≥-2. q<1000, so 12-2ⁿ⁺²<1000, 3-2ⁿ<250, 2ⁿ>-247, which is true for all n. So the range is -2≤n≤1, making n ∈ {-2 -1 0 1} giving (p,q,n)={ (13,11,-2) (14,10,-1) (16,8,0) (20,4,1) }.

For n=1, q=12-8=4 and p=12+8=20.

4×5/2=10, 20×21/2=210; 210-10=200=100×2.

Therefore 5+6+7+...+19+20=200.

Now let p+q+1=2ⁿ⁺³ and p-q=25. 

p+q=2ⁿ⁺³-1, p-q=25. Add these: 2p=2ⁿ⁺³+24, p=2ⁿ⁺²+12, q=2ⁿ⁺²-13.

q>0, so 2ⁿ⁺²>13, n+2≥4, n≥2. But p<1000, 2ⁿ⁺²+12<1000, 2ⁿ⁺²<988, n≤7.

Therefore, 2≤n≤7.

For n=7, p=524 and q=499.

524×525/2=137550 and 499×500/2=124750, 137550-124750=12800.

Therefore: 500+501+...+523+524=12800.

Continued in comment...

Comments

Note that there are many more solutions to this problem and all can be found by applying the method shown and using all values of n in the discovered range.

Also, (p-q)(p+q+1)=5×(5×2ⁿ⁺³) and solving for simple systems (p-q=5 and p+q=5×2ⁿ⁺³-1) and (p-q=5×2ⁿ⁺³ and p+q=4) produce more solutions for (p,q,n), including:

(642,637,5): 638+639+640+641+642=3200;

(7,2,-2): 3+4+5+6+7=25.